ZetaX

Do engineers really call them "infinities"¿ Sounds like a completely weird naming scheme... Note that there are several ways to turn "infinity" into a number (but normally, it should not be treated like a number at all, more like an exception akin to NaN), for example http://en.wikipedia.org/wiki/Surreal_number .

Mazon Del: that's not an infinity, but a degree of freedom. Something completely different.

I really want to keep it civil, but I would want to add a word for this to be possible: be rational and reasonable. If a argument, especially such a solid one as mathematical proof, is given, you can't just apply your version of wishful thinking. To make my point clear: I gave a proof in my second to last post (the same I already gave 3 times). Either accept it or point out errors; other things are not accepteble. This is mathematics, where decisions are not made by popular vote (not even a closed circle votes as in the dwarf planet demotion), but by the power of logical conclusions only.

Make me. Seriously, I will continue to use you as a bad example as long as you continue your behaviour. Also, the example I gave is very close to the error K^2 makes. You are also still invited to point out actual errors. You just choose not to, or you can't find any; "your argument is invalid" is not a proper argument, so make better ones in the future.

I was talking about the multipicative one, obviously. Please stop trying to "contradict" things that way, it is just tedious to repeat the whole terms every sentence. I did. I gave both examples and proofs. You instead gave only examples that are flawed (the 1 is an identity even for 0; and the linked wikipedia list has not a single example supporting what you say). Name a definition of a^b that does not simply extend to 0^0 without any problem. The map x^y is simply 1 at (0,0); what you wan to say is that it is not continuous there. But that does not make the definition (it is one) invalid. Then you are obviously not using a power series; so that's not an answer at all. Or you seriously think that x^0 will evaluate wrongly if x is close to 0, which it won't; at least if you set 0^0=1. Clearly it fails if you consider 0^x instead, but that's again just the lack of continuity (and smoothness), but even more, it is not the example I gave, thus: stop using strawmen. What. I used the definition of "identity" (look up any math page of wikipedia). I even repeated what that definition means. And I will repeat qemist's (he did by the way, like you, not state all his axioms in full, thus no, it is not as precise as mine; but I don't even claim his proof is wrong, just that it uses more properties than necessary) proof for you, and if you think there is an error or he did different: A: Assume we have a left neutral L: LA = a for all a. B: Assume we have a left neutral R: aR = a for all a. C: Assume the structure is associative. D: Assume we have left-inverses a-L (in regard to L): a-Ra = L for all a. E: Assume we have right-inverses a-R (in regard to R): aa-R = R F: Assume equality is transitive (I really don't think I need to mention this...). Idea: evaluate LR in two ways (I will supress mentioning associativity, and instead write no bracket at all): qemist's version: 1) LR = Laa-R = aa-R = R [using axioms D, A, D in that order] 2) LR = a-LaR = a-La = L [using axioms E, B, E in that order] Thus L = LR = R, so by F we have L=R. Now in comparision, my argument; poin out any presumed errors explicitely! 1) LR = R [use axiom A, with the special case of a = R] 2) LR = L [use axiom B, with the special case of a = L] Again get L = LR = R and conclude by F that L=R. Note that C,D,E were not used at all. I even only used A and B for a being the neutral elements. Also, it's your turn to at least give SOME definition. You act like mine is "wrong" despite it being the accepted one in mathematics, and without you ever demonstrating any problems with that. You just calculate as you want, sometimes following a rule/axiom and sometimes claiming one is not allowed to use it. That's definitely no how it works. You are, as gpisics behaviour alraedy demonstrates (*sigh*), on a road to an argument like "what if I can transmit information instantly", but ignore that relativity does make that "instantly" very observer dependent. Your construct needs to be self-consistent; or, and it's not even that; at least have a definition. You can change the rules, but you a) can't break logics itself, need to propery say what it means. You did neither. And my proof uses only that and the definition of "identity". Point out an error or stop acting that way. I never said every magma has one; I only said that the existence of an identity on each side implies they are equal; and then every other identity is equal to that, too. You claimed you can have distinct identities on both sides in a magma, and I proved you wrong; just accept it. I don't get what is so difficult to understand here. If the proof above still is no explicit enough, answer me the following: a) Give the definition of left and right identity. Are we at least using a magma¿ c) Do you accept that = is transitive¿ d) What is not clear or what do you consider faulty. In all seriousity, the last time I needed to expand down to that level for formality for such a triviality was when I explained groups to 10th graders not making this up; giving lectures to interested students all he time). I knon you are a physicist and would have expected a better mathematical understanding from such (despite the jokes about them).

That one is completely different. 1 is a multiplicative identitiy, even for 0. But even if not, that would not be a problem, as you would have defined 0 (additive identity) before you define 1 (multiplicative identity; only field without 0 should be a group); and we are talking about two structures here, not just one. As I said, an identity is one for all elements in question; that's the definition. All you could do is make a definition such as "a left-neutral element is one such that e°x=x for all x" and "a right-neutral is one such that x°e'=x for all x except left-neutrals". That would be a correct definition, but it would a) not be the accepted one, be asymetrical (calling the latter a "almost-right-neutral element" would make more sense, for example). Or you go with "a set of identities is a set S of elements such that s°x=x for all s in S and all x not in S". Then every single real number is an identity if you just wish so: set S = IR [the reals]; then clearly s°x=x for all x in IR-IR aka the empty set. That definition would indeed not be problematic, but it is again of a different type. And by the way, 0^0 = 1; that's not just some random convention, but the one accepted by mathematicians. It works very well, for example: a^b is the cardinality (i.e. the number/amount of elements) of the set A^B := {f:B->A} of maps f from a set B with b elements to a set A with a elements. And there is indeed exactly one map from the empty set to itself. And when was the last time you wrote a polynomial/power series not as sum a_i x^i, but as a_0 + sum a_i x^i instead¿ He still uses that a°e=a for a right identity e (similiar for left); the same reasoning I used. I used the defining propterty of identity; he used that, plus associativity plus inverses of some kind. Yet you accept his, which uses everything I use, but not mine, which works in a more general case by the very same argument. Do you really agree with his proof more because he wrote a (general element) instead of e (neutral one of some kind)¿ So where exactly do you disagree with both my proof and Wikipedia¿ That magma can only have distinct identities if they are all on the same side. unless you now also doubt transitivity of "=".

You really only need the definition of "identity" to conclude that left ones are equal to right ones if both exist. For e to be a left identity means to satisfy e°x=x for all x. For e' to be a right identity means to satisfy x°e'=x for all x. You can't just exclude this properties in the cases where x is an "identity"; how the heck would that definition even work¿ You would define identity using the word identity, making a circular definition. The list you gave has not a single such example, too; only cases where only identities on one side exist. It even says: "But if there is both a right identity and a left identity, then they are equal and there is just a single two-sided identity". Proof: If you have such e and e' then automatically e = e°e' = e', both equalities by definition of the respective identities. And if there is another one on any side, the same shows it's equal to e.

What logic¿ He just said that one "could conceive" such a thing and unlike your example, he still tried to adhere to the definition. The mathematical definition of 0 is to be the neutral element of addition and as such it is automatically its own inverse. If you conceive a new object, then you give it a new name; if it properly generalises the old one, then you can think about using the same name, but otherwise it would just be completely against convention and common sense. The argument would sound like someone answering my claim that the moon is not made of cheese by "but I named that cheese in my hand 'Moon', yo sou are wrong"; no, I am not, that's just a plain equivocation fallacy.

That graphical "proof" only talks about things that already contain the reals (probably you talk about a real vector space here). The concept of "0" exists in every abelian group, especially in every ring like e.g. IF_2 = IZ/2 = {0,1}, the field with two elements.

I never said it does. Why would I even mention non-unique inverses otherwise¿ Unless you just want to only have the two zeroes, you need to define how other elements are added. And what is 0_L + 0_R¿ If they actually are left and right identities, then it would be both 0_L and 0_R, making them equal. If you use "+" for the structure of a group/monoid/whatever, then it is a pretty well established convention that this implictely means it is abelian/commutative. If your structure lacks that property, you should use other symbols like ·, * or just "nothing". As a result, 0 will always be inverse (on both sides) to itself. Also note that "algebra" is already in use (essentially: a morphism of rings with a fixed source), "algebraic structure" is the better term for this. Then don't call it 0. Calling it 0 without it being the neutral element of "addition" is like calling your hamster "Zero": you can, but what is the point¿ Actually you can't: set theory needs several more axioms than just "the empty set exists" and the few axioms that construct new sets from old ones. The axiom of choice comes to mind, but even the simpler existence of the set omega must be explicitely required.

No, not really. The symbol 0 is defined to denote the neutral element of addition, and the reasoning you gave shows that it is its own inverse (or at least, _an_ inverse, there might be more if we don't have inverses or associativity). Thus you would need to break that very definition to make 0 different from -0.

Which won't work.

Not only EM, the nuclear forces will also have something to say.

1/4 What is "acoustic frequency electromagnetic field" even supposed to mean¿ All of your post sounds very weird.

You would need to define "center of mass" first. You obviously need to assume some finiteness condition on the universe (infinite mass will be a problem unless somehow spaced out thinner the far you go out), but you also need to to define what "the universe" even is: depending on your frame of reference, it will look differently. Also, even if fixing a frame of reference (which sounds very asymetrical) I don't see how to define centers of masses unless requiring some flatness condition on space. To maybe better understand some of the problems (and using the bad analogy of spacetime being like that): where on the earth's surface (!) is the center of mass of all humans¿

Even if accurate, the main problem would be that somehow suddenly one gas giant gets many times larger in mass. In reality, such a discontinuous effect makes almost no sense. The mass might e.g. be streaming in from outside the solar system over the course of many many years. Many such scenarios are less brutal to the existing orbits than just "make it bigger".

The standard power network with its 25-100Hz has a frequency much more closer to the brain (with some free interpretations what that even means it is somewhat around 10Hz if I remember correctly) and yet almost no one complains about that. The main reason by which other sources like mobile phones or wireless can have an effect is heating, and considering the inverse square law even a much stronger sender than your router won't suffice for that unless you use them as a pillow.

It's likely that this is possible, and probably not even difficult. The information itself is almost entirely consrved in the neural structure, and humans survive (mostly) to even have cut parts out or things like epilepsy/migraine that cause neurons to fire randomly.

What. That makes no sense. We even do special relativity (at least some aspects of it) in high school, and it is taught pretty early in university. And any course on general relativity (which will exist at any serious university) will require or recall special relativity. Do you even know what special relativity is¿ Yeah, ignore my post on symmetry of physical laws (which are very well tested by the way). Look up the Noether principle, if your claim is true you would not only break special and general relativity, but also conservation of momentum and energy.

What weird definition do you use for this to be possible¿

That user is me. I will just shorten two (of some more) main arguments for you (did not check if the first one was adressed before): a) If everything is deterministic, then so is the judical system, including us putting people to prison. The main point of a modern society's judical system is not to punish peiple (that would be archaic revenge-oriented thinking), but to give people a strong incentive to (not) do things. This is equally valid without consiousness.

What are those "enormous implications" of its absense then¿ There is probably not a single social thing that changes if we prove it to be nonexistent; there are probably more implications if it would actually exist.

Actually, I want to give that question right back at you. You just randomly claim there is a fundamental difference between brain research and cancer research that goes as deep as intervening with statistics itself.

What has this to do with the matter¿ N_las made a statistical statement, not a sociological one.

If you are just using a "very fast but slower than c"-drive, then yes. But that's not what the OP was talking about, he needs to catch those photons. Then the scenario in the OP would not even be possible. A true Alcubierre drive can go FTL, it's just not locally breaking c. And if you can go faster than c, then there is no need to "mangle" the message. Just fly around it: go 10 dgeree to the "left", fly a bit faster than originally needed, and then correct at the end, then wait for the photons to arrive.